1、求函数y=根号x^2+6x+25 +根号x^2-10x+29的值域
3个回答
1
y=√(x^2+6x+25) + √(x^2-10x+29)
=√[(x+3)^2+4^2]+ √[(x-5)^2+4]
A(-3,4) B(5,-2)
AB直线：y+2=[(-2-4)/(5+3)] (x-5)
y=0,x=-8/3+5=7/3
x=7/3
y最小=√[(7/3+3)^2+16] + √[(7/3-5)^2+4] =(4/3)*5+(1/3)*10=10
y≥10
2
y=√(x^2+6x+25)-√(x^2-10x+29)
=√[(x+3)^2+16 - √[(x-5)^2+4]
A1(-3,4) B1(5,2)
直线A1B1：y-2=[(2-4)/(5+3)] (x-5)
y=0,x=(-2)*(-4)+5=13
x=13时,y最大=√[(13+3)^2+16] - √[(13-5)^2+4]
=4√17 - √68
=2√17
y=√(x^2+6x+25)-√(x^2-10x+29)
=[(x^2+6x+25)-(x^2-10x+29)] / [√(x^2+6x+25)+√(x^2-10x+29)
lim(x→-∞)√(x^2+6x+25)-√(x^2-10x+29)
=lim(x→-∞)(16x-4)/[√(x^2+6x+25)+√(x^2-10x+29)]
=-16/2=-8
2√17≥√(x^2+6x+25)-√(x^2-10x+29) >-8
3
y=(x-a)^2-2|x|
=x^2-2ax+a^2-2|x|
=x^2-2x(a+|x|/x)x+a^2
=[x-(a+|a|/x)]^2+a^2-(a+|x|/x)^2
x0
a^2-(a+1)^2=-2a-1
因此a=0时,y=(x-a)^2-2|x|最小值-2a-1
